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Fundamentals of Robotic MechanicalSystems

Chapter 3: Fundamentals of Rigid-Body Mechanics

Jorge Angeles

Department of Mechanical Engineering &Centre for Intelligent Machines

McGill University

2009

Jorge Angeles (McGill University ) Ch3: Fundamentals of Rigid-Body Mechanics 2009 1 / 66

Overview

IntroductionGeneral Rigid-Body Motion and Its Associated Screw

I The Screw of a Rigid-Body MotionI The Plücker Coordinates of a LineI The Pose of a Rigid Body

Rotation of a Rigid Body About a Fixed PointGeneral Instantaneous Motion of a Rigid Body

I The Instant Screw of a Rigid-Body MotionI The Twist of a Rigid Body

Acceleration Analysis of Rigid-Body MotionsRigid-Body Motion Referred to Moving Coordinate AxesStatic Analysis of Rigid BodiesDynamics of Rigid Bodies


General Rigid-Body Motion and Its Associated Screw

Point A : a reference point of rigid body B with the position vector a in theoriginal configuration and a′ in the displaced configuration, A′.

Point P : an arbitrary point of B with the position vector p in the originalconfiguration and p′ in the displaced configuration, P ′.

Determine p′ with given a, a′, p, and the rotation matrix Q :

p′ − a′ = Q(p− a) ⇒ p′ = a′ + Q(p− a) (3.2)


General Rigid-Body Motion and Its Associated Screw(Cont’d)

Displacements of A and P :

dA ≡ a′ − a, dP ≡ p′ − p (3.3)

From eqs.(3.2) and (3.3) :

dP = p′ − p = a′ − p + Q(p− a)= a′ − a− p + Q(p− a) + a

⇒ dP = dA + (Q− 1)(p− a) (3.5)



Theorem

The component of the displacements of all the points of a rigid bodyundergoing a general motion along the axis of the underlying rotation is aconstant.

Proof : Multiply both sides of eq.(3.5) by eT :

eTdP = eTdA + eT (Q− 1)(p− a)

Qe = e → eTQ− eT = 0T → eT (Q− 1) = 0T

⇒ eTdP = eTdA ≡ d0 (3.6)



Theorem (Mozzi, 1763 ; Chasles, 1830)Given a rigid body undergoing a general motion, a set of its points located ona line L undergo identical displacements of minimum magnitude. Moreover,line L and the minimum-magnitude displacement are parallel to the axis ofthe rotation involved.

Proof : dP = d‖ + d⊥ (3.7a)where d‖ = eeTdP = d0e, d⊥ = (1− eeT )dP (3.7b)

‖dP ‖2 = ‖d‖‖2 + ‖d⊥‖2 = d20 + ‖d⊥‖2



Minimize displacement ‖dP ‖ : d⊥ = 0⇒ dP = αe

⇒ All points of minimum-magnitude displacement lie in a line parallel to the axisof rotation of Q, q.e.d.

If P ∗ is a point of minimum-magnitude displacement, then

d∗⊥ ≡ (1− eeT )dP∗ = (1− eeT )dA + (1− eeT )(Q− 1)(p∗ − a) = 0

⇒ (1− eeT )dA + (Q− 1)(p∗ − a) = 0

Above equation holds if αe is added to p∗,⇒ Screw motion of axes L and pitch p :

p ≡ d0

φ=

dTP eφ

or p ≡ 2πd0

φ(3.8)


The Screw of a Rigid-Body Motion

The screw axis L : specified by P0 and e.If P0 is closest to the origin, then

eTp0 = 0 (3.9)

and its displacement d0 is identical to d‖ :

(Q− 1)d0 = 0

where, from eq.(3.5),

d0 = dA + (Q− 1)(p0 − a) (3.10a)

and, from eq.(3.7b),

dA + (Q− 1)(p0 − a) = d‖ ≡ eeTd0


The Screw of a Rigid-Body Motion(Cont’d)

From Theorem 1

eTd0 = eTdA ⇒ dA + (Q− 1)(p0 − a) = eeTdA

or(Q− 1)p0 = (Q− 1)a− (1− eeT )dA (3.10b)

Adjoin eqs.(3.9) and (3.10b) :

Ap0 = b (3.11)

whereA ≡

[Q− 1

eT

], b ≡

[(Q− 1)a− (1− eeT )dA

0

](3.12)



ATAp0 = ATb (3.13)

Substituting into eq.(3.12) expression

Q = eeT + cosφ(1− eeT ) + sinφE

yields ATA = 2(1− cosφ)1− (1− 2 cosφ)eeT

or (ATA)−1 = α1 + βeeT .

Since (ATA)(ATA)−1 = 1,

α =1

2(1− cosφ), β =

1− 2 cosφ2(1− cosφ)

(3.16)



⇒ (ATA)−1 =1

2(1− cosφ)1 +

1− 2 cosφ2(1− cosφ)

eeT (3.17)

On the other hand,

ATb = (Q− 1)T [(Q− 1)a− dA] (3.18)

Upon solving eq.(3.13) for p0 and substituting relations (3.17) and (3.18)into the expression thus resulting, one finally obtains

p0 =(Q− 1)T (Qa− a′)

2(1− cosφ), for φ 6= 0 (3.19)


The Plücker Coordinates of a Line

FIGURE: A line L passing through two points.

(p2 − p1)× (p− p1) = 0

⇒ (p2 − p1)× p + p1 × (p2 − p1) = 0


The Plücker Coordinates of a Line(Cont’d)

(P2 −P1)p + p1 × (p2 − p1) = 0

[P2 −P1 p1 × (p2 − p1)

] [p1

]= 0 (3.21)

We define

γL ≡[

p2 − p1

p1 × (p2 − p1)

](3.22)

Components of γL : Plücker coordinates of L.



Lete ≡ p2 − p1

‖p2 − p1‖, n ≡ p1 × e (3.23)

⇒ γL = ‖p2 − p1‖[en

]Vector e determines the direction of L, while n, the moment of L,

determines its location.Plücker array κ :

κ =[en

](3.24)



There are only four independent Plücker coordinates of line L since :

e · e = 1, n · e = 0 (3.25)

That is,

(i) the sum of the squares of the first threecomponents of a Plücker array is unity, and

(ii) the unit vector of a line is normal to the momentof the line

⇒ Set of Plücker arrays does not form a vector space



nA ≡ (p− a)× e, nB ≡ (p− b)× e (3.26)

⇒ κA ≡[

enA

], κB ≡

[e

nB

](3.27)

nB − nA = (a− b)× e (3.28)

⇒ κB =[

enA + (a− b)× e

](3.29)

or κB = UκA : Plücker-coordinate transfer formula

U ≡[

1 OA−B 1

](3.31)



det(U) = 1

⇒ U belongs to the unimodular group of 6× 6 matrices

⇒ U−1 =[

1 OB−A 1

](3.33)

Note that from eq.(3.28) :

(a− b)TnB = (a− b)TnA (3.34)

⇒ The moments of the same line L with respect to two points are notindependent : they have the same component along the line joining thetwo points.



A line at infinity : the orientation is undefined but a direction of its moment isdefined

If we rewrite eq.(3.24) in the form

κ = ‖n‖[e/‖n‖n/‖n‖

], f ≡ n/‖n‖

then, lim‖n‖→∞

κ =(

lim‖n‖→∞

‖n‖)(

lim‖n‖→∞

[e/‖n‖

f

])

⇒ lim‖n‖→∞

κ =(

lim‖n‖→∞

‖n‖)[

0f

]⇒ κ∞ =

[0f

](3.35)


The Pose of a Rigid Body

A rigid-body motion is fully described by six independent or seven dependentparameters.Hence, if a rigid body undergoes a general motion of rotation Q anddisplacement d from a reference configuration C0, then pose array, or thepose, s of the body, is defined as a 7-dimensional array :

s ≡[qT q0 dTA

]T (3.36)

where q and q0 are any four invariants of Q.


The Pose of a Rigid Body (Cont’d)

For example, Euler-Rodrigues parameters :

q ≡ sin(φ/2)e, q0 ≡ cos(φ/2)

linear invariants : q ≡ (sinφ)e, q0 ≡ cosφ

with ‖q‖2 + q20 = 1 ;

the natural invariants : q ≡ e, q0 ≡ φ with ‖e‖2 = 1.

Pose estimation : the computation of the attitude of a rigid body, given bymatrix Q or its invariants from coordinate measurements over a certain finiteset of points.



FIGURE: Helmet-mounted display system (CAE Electronics, St.-Laurent, Quebec, Canada.)



FIGURE: Decomposition of the displacement of a rigid body.

To determine matrix Q, define P1, P2, and P3 so that

e1 ≡−−→AP1, e2 ≡

−−→AP2, e3 ≡

−−→AP3

ei · ej = δij , i, j = 1, 2, 3, e3 = e1 × e2

where δij is the Kronecker delta, defined as 1 if i = j and 0 otherwise.



Let qij be the entries of Q in X, Y, Z with the origin at A and with X, Y, Zparallel to e1, e2, e3. From Definition 2.2.1, qij = ei · e′j , i.e.,

[ Q ] =

e1 · e′1 e1 · e′2 e1 · e′3e2 · e′1 e2 · e′2 e2 · e′3e3 · e′1 e3 · e′2 e3 · e′3

(3.42)

Note that all ei and e′i in eq.(3.42) must be represented in the samecoordinate frame.

Then, use, for example, eq.(3.19) to determine the point of the screw axis thatlies closest to the origin and compute the Plücker coordinates of the screwaxis.


Rotation of a Rigid Body About a Fixed Point

Fully described by a rotation matrix Q that is proper orthogonal and is asmooth function of time. Then,

p(t) = Q(t)p0 (3.43)

⇒ p(t) = Q(t)p0 (3.44)

Solve eq.(3.43) for p0 and substitute the result into eq.(3.44) :

p = QQTp (3.45)

The angular-velocity matrix :

Ω ≡ QQT (3.46)


Rotation of a Rigid Body About a Fixed Point (Cont’d)

TheoremThe angular-velocity matrix is skew symmetric.

The angular-velocity vector ω of the rigid-body motion is defined as

ω ≡ vect(Ω) (3.47)

and hence, eq.(3.45) can be written as

p = Ωp = ω × p (3.48)

⇒ The velocity of any point P of a body moving with a point O fixed isperpendicular to line OP .


General Instantaneous Motion of a Rigid Body (Cont’d)

p(t) = a(t) + Q(t)(p0 − a0) (3.49)

⇒ p(t) = a(t) + Q(t)(p0 − a0) (3.50)

Then, eq.(3.49) is solved for (p0 − a0) and the result is substituted intoeq.(3.50) :

p = a + Ω(p− a) (3.51)

⇒ p = a + ω × (p− a) (3.52)

⇒ (p− a) · (p− a) = 0 (3.53)

TheoremThe relative velocity of two points of the same rigid body is perpendicular tothe line joining them.


General Instantaneous Motion of a Rigid Body (Cont’d)Upon dot-multiplying both sides of eq.(3.52) by ω :

ω · p = ω · a = constant ⇒

CorollaryThe projections of the velocities of all the points of a rigid body onto theangular-velocity vector are identical.

TheoremGiven a rigid body under general motion, a set of its points located on a line L′undergoes the identical minimum-magnitude velocity v0 parallel to the angularvelocity.

DefinitionThe line containing the points of a rigid body undergoing minimum-magnitudevelocities is called the instant screw axis (ISA) of the body under the given motion.


The Instant Screw of a Rigid-Body Motion

The instantaneous motion of a body is equivalent to that of the bolt of a screwof instantaneous axis L′, the ISA, and is called an instantaneous screw.Since v0 ‖ ω,

v0 = v0ω

‖ω‖(3.54)

⇒ the pitch: p′ ≡ v0‖ω‖

≡ p · ω‖ω‖2

or p′ ≡ 2πv0‖ω‖

(3.55)

Plücker coordinates of ISA L′ : pL′ ≡[e′

n′

](3.56)

with e′ ≡ ω

‖ω‖, n′ ≡ p× e′ (3.57)


The Instant Screw of a Rigid-Body Motion (Cont’d)

An instantaneous rigid-body motion is defined by a line L′, a pitch p′, and anamplitude ‖ω‖ :

−∞ ≤ p′ ≤ +∞

A line supplied with a pitch : a screw

A screw supplied with an amplitude representing the magnitude of an angularvelocity : a twist.



The ISA can be alternatively described in terms of p′0, its point lying closestto the origin. Let

a ≡ a‖ + a⊥ (3.58)

wherea‖ ≡ a · ω ω

‖ω‖2≡ ωωT

‖ω‖2a,

a⊥ ≡(

1− ωωT

‖ω‖2

)a ≡ − 1

‖ω‖2Ω2a (3.59)

if we use the identity in (2.39) :Ω2 ≡ ωωT − ‖ω‖21, (3.60)



Substitute eq.(3.59) into eq.(3.52) :

p =ωωT

‖ω‖2a︸︷︷︸

p‖

− 1‖ω‖2

Ω2a + Ω(p− a)︸︷︷︸p⊥

(3.61)

ωωT

‖ω‖2 a : the axial component of p (‖ to ω)

− 1‖ω‖2 Ω2a + Ω(p− a) : the normal component of p (⊥ to ω)

Is it possible, for an arbitrary motion, to find a certain point of position vectorp whose velocity normal component vanishes ?



p⊥ = 0 ⇒ Ω(p− a)− 1‖ω‖2

Ω2 a = 0 (3.62)

⇒ Ωp = Ω(

a +1‖ω‖2

Ωa)

(3.63)

orΩ(p− r) = 0 (3.64a)

withr ≡ a +

1‖ω‖2

Ωa (3.64b)



A possible solution is

p = r = a +1‖ω‖2

Ωa (3.65)

This solution is not unique :

αω + p also satisfies eq.(3.63)

⇒ eq.(3.63) determines a set of points lying on the ISA.



For example, we can find the point of the ISA p′0 lying closest to the origin.Obviously :

ωTp′0 = 0 (3.66)

If eq.(3.63) is rewritten for p′0 and eq.(3.66) is adjoined to it, then,

Ap′0 = b (3.67)

whereA ≡

[ΩωT

], b ≡

[Ωa + (1/‖ω‖2)Ω2a

0

](3.68)



Multiply eq.(3.67) by AT : ATAp′0 = ATb (3.69)

⇒ ATA = ΩTΩ + ωωT = −Ω2 + ωωT (3.70)

Use eq.(3.60) :Ω2 ≡ ωωT − ‖ω‖21

Hence, we have

ATA = ‖ω‖21, ATb = Ω(a−Ωa) (3.71)



Substitute eqs.(3.71) into eq.(3.69) and solve for p′0 :

p′0 =Ω(a−Ωa)‖ω‖2

≡ ω × (a− ω × a)‖ω‖2

(3.72)

⇒ the instantaneous screw is fully defined by an alternative set of 6independent scalars : 3 components of ω and 3 components of a.


The Twist of a Rigid Body

A pitch p added as a fifth feature to the Plücker array of a line⇒ a screw s :

s ≡[

ep× e + pe

](3.73)

An amplitude A multiplying the screw produces a twist or a wrench as an8-parameter array : the amplitude, the pitch, and the 6 Plücker coordinates ofthe associated line, with only 6 of them independent.

A twist defines completely the velocity field of a rigid body : 3 componentsof the angular velocity and 3 components of the velocity of any of the pointsof the body.


The Twist of a Rigid Body (Cont’d)

Upon multiplication of the screw in eq.(3.73) by A :

t ≡[

Aep× (Ae) + p(Ae)

]Here : Ae is the angular velocity ω ‖ e of magnitude A.

The lower vector of t, v, is the velocity of point O of a rigid body movingwith an angular velocity ω so that point P moves with a velocity pω ‖ ω :

v = −ω × p + pω

⇒ the twist t ≡[ωv

](3.74)



The screw of infinitely large pitch :

limp→∞

[e

p× e + pe

]≡ lim

p→∞

(p

[e/p

(p× e)/p+ e

])=

(limp→∞

p

)[0e

]

The screw of infinite pitch s∞ ≡[0e

](3.75)

is identical to the Plücker array of the line at∞ in a plane of unit normal e.



The relationships between ω and the time derivatives of the invariants of theassociated rotation :

ν ≡[eφ

], λ ≡

[(sinφ)e

cosφ

], η ≡

[[sin(φ/2)]ecos(φ/2)

],

ν = Nω, λ = Lω, η = Hω (3.77a)

N ≡[[sinφ/(2(1− cosφ))](1− eeT )− (1/2)E

eT

],

L ≡[(1/2)[tr(Q)1−Q]−(sinφ)eT

], H ≡ 1

2

[cos(φ/2)1− sin(φ/2)E

− sin(φ/2)eT

]



The inverse relations of (3.77a) :

ω = Nν = Lλ = Hη (3.78a)

where

N ≡[(sinφ)1 + (1− cosφ)E e

],

L ≡[1 + sinφ

1+cosφE − sinφ1+cosφe

],

H ≡ 2[[cos(φ/2)]1 + [sin(φ/2)]E −[sin(φ/2)]e

]Caveat : The angular velocity vector is not a time-derivative, i.e., no

Cartesian vector exists whose time-derivative is the angular-velocity vector.



The relationships between the twist and the time-rate of change of the7-dimensional pose array s :

s = Tt (3.79)

where

T ≡[F O43

O 1

](3.80)

andt = Ss (3.81a)

where

S ≡[

F OO34 1

](3.81b)



The relation between the twists of the same rigid body at two different points, A andP :

tA =[ωvA

], tP =

[ωvP

](3.82)

Equation (3.52), rewritten as vP = vA + (a− p)× ω, combined with eq.(3.82)yields

tP = UtA where U ≡[

1 OA−P 1

](3.84)

where A and P denote the cross-product matrices of vectors a and p, respectively.Equation (3.84) is called the twist-transfer formula.


Acceleration Analysis of Rigid-Body Motions (Cont’d)

Differentiate eq.(3.51) with respect to time :

p = a + Ω(p− a) + Ω(p− a) (3.86)

Solving eq.(3.51) for p− a and substituting the result into eq.(3.86), weobtain

p = a + (Ω + Ω2)(p− a) (3.87)

whereW ≡ Ω + Ω2 (3.88)

is termed the angular-acceleration matrix.



The first term of the right-hand side of eq.(3.88) is skew-symmetric, thesecond one is symmetric. Thus,

vect(W) = vect(Ω) = ω (3.89)

the angular-acceleration vector. Moreover,

tr(W) = tr(Ω2) = tr(−‖ω‖21 + ωωT )= −‖ω‖2tr(1) + ω · ω = −2‖ω‖2 (3.90)

Equation (3.87) can be written as

p = a + ω × (p− a) + ω × [ω × (p− a)] (3.91)



The twist rate : t ≡[ωv

](3.92)

Differentiate eq.(3.79) : s = Tt + Tt

T ≡[F O43

O O

](3.94)

and F is one of N, L, or H, accordingly. For the inverse relationship,differentiate (3.81a) :

t = Ss + Ss where S =

[˙F O

O34 O

](3.96)



Let

λ ≡[

uu0

], η ≡

[rr0

](3.97a)

where

u ≡ sinφe, u0 ≡ cosφ,

r ≡ sin(φ

2

)e, r0 ≡ cos

(φ

2

)



N =1

4(1− cosφ)

[Be

]H =

12

[r01− R−rT

]

L =12

[1tr(Q)− Q

−ωT tr(Q) + ωT QT

]= −1

2

[2(ω · u)1 + ΩQωT tr(Q)− ωT QT

]where R denotes the cross-product matrix of r,

B ≡ −2(e · ω)1 + 2(3− cosφ)(e · ω)eeT − 2(1 + sinφ)ωeT

−(2 cosφ+ sinφ)eωT − (sinφ)[Ω− (e · ω)E]

andtr(Q) ≡ tr(ΩQ) ≡ −2ωT u



We also have :

˙N =[φ(cosφ)1 + φ(sin θ)E e

]˙L =

[V/D u

] ˙H =[r01 + R −r

]where

V ≡ U− (uuT + uuT )− u0

D(U− uuT )

D ≡ 1 + u0

with U denoting the cross-product matrix of u.


Rigid-Body Motion Referred to MovingCoordinate Axes

FIGURE: Fixed and moving coordinate frames.

F : the fixed coordinate frame X, Y, ZM : the moving coordinate frame X , Y, Z


Rigid-Body Motion Referred to MovingCoordinate Axes (Cont’d)

[ p ]F = [ o ]F + [ρ ]F (3.100)

[ρ ]F = [ Q ]F [ρ ]M (3.101)

Substitute eq.(3.101) into eq.(3.100) :[ p ]F = [ o ]F + [ Q ]F [ρ ]M (3.102)

⇒ [ p ]F = [ o ]F + [ Q ]F [ρ ]M + [ Q ]F [ ρ ]M (3.103)

From the definition of Ω in eq.(3.46) :

[ Q ]F = [ Ω ]F [ Q ]F (3.104)


Rigid-Body Motion Referred to Moving Coordinate Axes(Cont’d)

Hence,[ p ]F = [ o ]F + [ Ω ]F [ Q ]F [ρ ]M + [ Q ]F [ ρ ]M (3.105)

Differentiate eq.(3.105) with respect to time :

[ p ]F = [ o ]F + [ Ω ]F [ Q ]F [ρ ]M + [ Ω ]F [ Q ]F [ρ ]M+[ Ω ]F [ Q ]F [ρ ]M + [ Q ]F [ ρ ]M + [ Q ]F [ ρ ]M

Substitute (3.104) into the above equation :

[ p ]F = [ o ]F + ( [ Ω ]F + [ Ω2 ]F )[ Q ]F [ρ ]M+2[Ω ]F [ Q ]F [ ρ ]M + [ Q ]F [ ρ ]M


Static Analysis of Rigid Bodies

FIGURE: Equivalent systems of force and moment acting on a rigid body.

nP = nA + (a− p)× f (3.108)

nP = nA + f × (p− a) (3.109)


Static Analysis of Rigid Bodies (Cont’d)

Similar to Theorem 3 :

TheoremFor a given system of forces and moments acting on a rigid body, if theresultant force is applied at any point of a particular line L′′, then theresultant moment is of minimum magnitude. Moreover, thatminimum-magnitude moment is parallel to the resultant force.

The minimum-magnitude moment n0 :

n0 = n0f‖f‖

, n0 ≡nP · f‖f‖

(3.110)



The pitch of the wrench, p′′ :

p′′ ≡ n0

‖f‖=

nP · f‖f‖2

or p′′ =2πnP · f‖f‖2

(3.111)

The Plücker array of the wrench axis :

pL′′ ≡[e′′

n′′

], e′′ =

f‖f‖

, n′′ = p× e′′ (3.112)

The wrench axis is fully specified, then, by f and P ′′0 lying closest to theorigin

p′′0 =1‖f‖2

f × (nA − f × a) (3.113)



Similar to Theorem 1 :

TheoremThe projection of the resultant moment of a system of moments and forcesacting on a rigid body that arises when the resultant force is applied at anarbitrary point of the body onto the wrench axis is constant.

To derive the wrench array w, we need to multiply the screw of eq.(3.73) byan amplitude A with units of force.We shall be able to obtain the power developed by the wrench on the bodymoving with the twist t but the inner product of these two arrays would bemeaningless.



We redefine the wrench as a linear transformation of s :

Γ ≡[O 11 O

]⇒ w ≡ AΓs ≡

[p× (Ae) + p(Ae)

Ae

]The first 3 components of w represent the moment of a force of magnitude A acting

along e with respect to P plus a moment parallel to e and of magnitude pA. The last3 components pertain to a force of magnitude A and parallel to e :

f ≡ Ae, n ≡ p× f + pf

⇒ w ≡[nf

]and Π = tT w (3.116)



Let

w = WΓsw , t = T st (3.117)

The two screws sw and st are reciprocal if

(Γsw)T st ≡ sTwΓT st = 0 (3.118)

or

sTwΓst = 0 and sTt Γsw = 0 (3.119)

For arbitrary points A and P

wA ≡[nAf

], wP ≡

[nPf

](3.120)



⇒ wP = VwA where V ≡[1 A−P0 1

](3.121)

The last relation is termed the wrench-transfer formula.

Multiplying the transpose of eq.(3.84) by eq.(3.121) yields

tTPwP = tTAUTVwA (3.123)

where

UTV =[1 −A + P0 1

] [1 A−P0 1

]= 16×6 (3.124)

Thus, tTPwP = tTAwA. Also note that V−1 = UT .


Dynamics of Rigid Bodies (Cont’d)

The rigid body mass :m =

∫BρdB (3.126)

where ρ is a mass density and B denotes the region occupied by the body.The mass first moment qO ≡

∫BρpdB (3.127)

The mass second moment (symmetric matrix)IO ≡

∫Bρ[ (p · p)1− ppT ]dB (3.128)

is called the moment-of-inertia matrix of the body with respect to O.

TheoremThe moment of inertia of a rigid body with respect to a point O is positivedefinite.



Proof : For any vector ω, the quadratic form ωT IOω is positive :

ωT IOω =∫Bρ[ ‖p‖2‖ω‖2 − (p · ω)2 ]dB (3.129)

Substituting p · ω = ‖p‖‖ω‖ cos(p,ω) (3.130)

into eq.(3.129) leads to :

ωT IOω =∫Bρ‖p‖2‖ω‖2[ 1− cos2(p,ω) ]dB

=∫Bρ‖p‖2‖ω‖2 sin2(p,ω)dB



The kinetic energy of a rigid body : T ≡∫B

12ρ‖p‖2dB

where p = ω × p = −Pω

⇒ ‖p‖2 = (Pω)TPω = ωTPTPω = −ωTP2ω

= ωT (‖p‖21− ppT )ω

Hence,T =

12

∫BρωT (‖p‖21− ppT )ωdB

=12ωT[∫Bρ(‖p‖21− ppT )dB

]ω

⇒ T =12ωT IOω (3.134)



The mass center of a rigid body is defined as a point C of position vector c :

c ≡ qOm

(3.135)

The centroidal mass moment of inertia :

⇒ IC ≡∫Bρ[ ‖r‖21− rrT ]dB (3.136)

where r ≡ p− c

The Theorem of Parallel Axes :

IO = IC +m(‖c‖21− ccT ) (3.138)



For a body acted upon by a wrench of force f applied at its mass center, and amoment nC ,the Newton equation : f = mc (3.139a)

the Euler equation : nC = ICω + ω × ICω (3.139b)

The momentum m and the angular momentum hC :m ≡ mc, hC ≡ ICω (3.140)

Moreover,m = mc, hC = ICω + ω × ICω (3.141)

⇒ f = m, nC = hC (3.142)



The inertia dyad :

M ≡[IC OO m1

](3.143)

Now the Newton-Euler equations can be written as

Mt + WMt = w (3.144)

whereW ≡

[Ω OO O

](3.145)

is termed the angular-velocity dyad. Note that

Wt = 0 (3.146)



The momentum screw about the rigid body mass center

µ ≡[ICωmc

]= Mt (3.147)

⇒ µ = Mt + Wµ = Mt + WMt (3.148)

The kinetic energy

T =12m‖c‖2 +

12ωT ICω =

12tTMt (3.150)

The Newton-Euler equations

µ = w (3.151)


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